Question

I worked the probability problem I just need someone to check the answer please!
In a batch of 50 batteries, four are defective. If we select 3 batteries, what is the probability that at least 2 are good.

Answer 1

I did P=1-P(0 good)P(1good)
my answer is 99%

Answer 2

4 out of 50 are bad, so 46 out of 50 are good

can we do a binomial approach?

Answer 3

\[\binom{50}{2}(\frac{46}{50})^2(\frac{4}{50})^{48}+\binom{50}{3}(\frac{46}{50})^{3}(\frac{4}{50})^{47}\]

Answer 4

i think 50s are 3 in the binomial code ...

Answer 5

a binomial approach is used when looking for the probability of an exact number of things.
i used the multiplication rule( that is what is used in the book to find at least prob.)

Answer 6

\[\binom{3}{2}(\frac{46}{50})^2(\frac{4}{50})^{1}+\binom{3}{3}(\frac{46}{50})^{3}(\frac{4}{50})^{0 }\]

Answer 7

exactly 2 or exactly 3 are at least 2

Answer 8

P(at least 2 good)=1-69/4900*1/4900
=.999999
=99%

Answer 9

P( 0 good) = ((4c3)(46c0))/(50c3)
=1/4900

Answer 10

it is a high number ... im getting 98.2 or so

Answer 11

P(1 good)= ((4c2)(46c1))/(50c3)
=69/4900

Answer 12

shold we assume replacement or not?

ggb = (4*46*47)/(50*49*48)
gbg = (4*46*47)/(50*49*48)
bgg = (4*46*47)/(50*49*48)
ggg = (46*47*48)/(50*49*48)

Answer 13

no replacement

Answer 14

3*(4*46*47)/(50*49*48)+(46*45*44)/(50*49*48)

without replacement gets us about 99.51

Answer 15

So in other words 99% would be a correct answer

Answer 16

im not the one grading it, but it seems fair enough :)

Answer 17

Thanks!

Answer 18

youre welcome